Average Error: 0.5 → 0.4
Time: 13.3s
Precision: binary64
\[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]
↓
\[\begin{array}{l} t_0 := \left(2 \cdot \pi\right) \cdot n\\ {t_0}^{\left(k \cdot -0.5\right)} \cdot \frac{{t_0}^{0.5}}{\sqrt{k}} \end{array} \]
\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)}
↓
\begin{array}{l}
t_0 := \left(2 \cdot \pi\right) \cdot n\\
{t_0}^{\left(k \cdot -0.5\right)} \cdot \frac{{t_0}^{0.5}}{\sqrt{k}}
\end{array}

(FPCore (k n)
 :precision binary64
 (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))
↓
(FPCore (k n)
 :precision binary64
 (let* ((t_0 (* (* 2.0 PI) n)))
   (* (pow t_0 (* k -0.5)) (/ (pow t_0 0.5) (sqrt k)))))
double code(double k, double n) {
	return (1.0 / sqrt(k)) * pow(((2.0 * ((double) M_PI)) * n), ((1.0 - k) / 2.0));
}
↓
double code(double k, double n) {
	double t_0 = (2.0 * ((double) M_PI)) * n;
	return pow(t_0, (k * -0.5)) * (pow(t_0, 0.5) / sqrt(k));
}

Error

Bits error versus k

Bits error versus n

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Derivation

  1. Initial program 0.5

    \[\frac{1}{\sqrt{k}} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\frac{1 - k}{2}\right)} \]

  2. Simplified0.4

    \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}}{\sqrt{k}}} \]

  3. Applied *-un-lft-identity_binary640.4

    \[\leadsto \frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(\mathsf{fma}\left(k, -0.5, 0.5\right)\right)}}{\color{blue}{1 \cdot \sqrt{k}}} \]

  4. Applied fma-udef_binary640.4

    \[\leadsto \frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\color{blue}{\left(k \cdot -0.5 + 0.5\right)}}}{1 \cdot \sqrt{k}} \]

  5. Applied unpow-prod-up_binary640.4

    \[\leadsto \frac{\color{blue}{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(k \cdot -0.5\right)} \cdot {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{0.5}}}{1 \cdot \sqrt{k}} \]

  6. Applied times-frac_binary640.4

    \[\leadsto \color{blue}{\frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(k \cdot -0.5\right)}}{1} \cdot \frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{0.5}}{\sqrt{k}}} \]

  7. Final simplification0.4

    \[\leadsto {\left(\left(2 \cdot \pi\right) \cdot n\right)}^{\left(k \cdot -0.5\right)} \cdot \frac{{\left(\left(2 \cdot \pi\right) \cdot n\right)}^{0.5}}{\sqrt{k}} \]

Reproduce

herbie shell --seed 2022081 
(FPCore (k n)
  :name "Migdal et al, Equation (51)"
  :precision binary64
  (* (/ 1.0 (sqrt k)) (pow (* (* 2.0 PI) n) (/ (- 1.0 k) 2.0))))